(2x^2+x)/3x-9=0

Solution for (2x^2+x)/3x-9=0 equation:

(2x^2+x)/3x-9=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We multiply all the terms by the denominator


(2x^2+x)-9*3x=0

Wy multiply elements

(2x^2+x)-27x=0

We get rid of parentheses

2x^2+x-27x=0

We add all the numbers together, and all the variables

2x^2-26x=0

a = 2; b = -26; c = 0;
Δ = b2-4ac
Δ = -262-4·2·0
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-26}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+26}{2*2}=\frac{52}{4}
=13
$